Question: $\sum\limits_{n=0}^{\infty }{{\left( -1 \right)}^{n+1}}\frac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}~~$ is the Maclaurin series for which function? Choose 1 answer: Choose 1 answer: (Choice A) A $\sin x$ (Choice B) B $-\sin x$ (Choice C) C $\cos x$ (Choice D) D $\cos (-x)$ (Choice E) E $-\cos x$
Answer: The given series has terms that alternate in sign and have only odd factorials in the denominator. Thus the given series closely resembles the Maclaurin series for the function $\sin x.$ $\sin x=x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}+...$ If we change every sign in the Maclauin series for $~\sin x\,$, we obtain the given series. Hence, $-\sin x=-x+\frac{{{x}^{3}}}{3!}-\frac{{{x}^{5}}}{5!}-...+{{\left( -1 \right)}^{n+1}}\frac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}+...$